= Solution
{c}
For the shifted normal proposal, $w(x)=e^{-\mu x+\mu^2/2}$ and the estimator is
$$
\hat p_N=\frac1N\sum_i1_{\{X_i>z\}}e^{-\mu X_i+\mu^2/2},
\qquad X_i\sim N(\mu,1).
$$
Let $\bar\Phi=1-\Phi$. Completing the square gives its unaveraged second moment
$$
I(\mu)=\int_z^\infty\frac{p(x)^2}{q_\mu(x)}\,dx
=e^{\mu^2}\bar\Phi(z+\mu).
$$
The variance is $[I(\mu)-\bar\Phi(z)^2]/N$, so only $I$ needs minimization. It tends to infinity at both ends of the real $\mu$ axis and has a finite minimizer. Differentiation at a minimizer yields
$$
2\mu\,\bar\Phi(z+\mu)=\varphi(z+\mu).
$$
For a standard normal variable, integration of $x\varphi(x)=-\varphi'(x)$ shows $E_p[X\mid X>a]=\varphi(a)/\bar\Phi(a)$. Consequently
$$
\boxed{2\mu=E_p[X\mid X>z+\mu].}
$$
The shift is chosen to reduce the second moment of the weighted rare-event indicator, rather than merely to increase the proposal's tail probability.
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